Let's break down the solution into **2 steps** with explanations at the end of each step, and then provide the final answer. --- ### **Step 1: Take the Laplace Transform of both sides** Given the differential equation and initial conditions: \[ y'' + 16y = 64t, \quad y(0) = 4, \quad y'(0) = 5 \] Take the Laplace transform of both sides (using properties of Laplace transforms for derivatives): - Laplace of \( y'' \) is \( s^2 Y(s) - s y(0) - y'(0) \) - Laplace of \( 16y \) is \( 16Y(s) \) - Laplace of \( 64t \) is \( 64 \cdot \frac{1}{s^2} \) So, \[ \mathcal{L}\{y''\} + 16\mathcal{L}\{y\} = \mathcal{L}\{64t\} \] \[ [s^2 Y(s) - s y(0) - y'(0)] + 16Y(s) = \frac{64}{s^2} \] Plug in initial values \( y(0) = 4 \), \( y'(0) = 5 \): \[ s^2 Y(s) - 4s - 5 + 16Y(s) = \frac{64}{s^2} \] **Explanation:** In this step, we've converted the original differential equation into an algebraic equation in terms of \( Y(s) \) using Laplace transforms and substituted the initial conditions. --- ### **Step 2: Solve for \( Y(s) \) and find the Inverse Laplace Transform** First, solve for \( Y(s) \): \[ s^2 Y(s) + 16Y(s) = \frac{64}{s^2} + 4s + 5 \] \[ Y(s)(s^2 + 16) = \frac{64}{s^2} + 4s + 5 \] \[ Y(s) = \frac{\frac{64}{s^2} + 4s + 5}{s^2 + 16} \] Simplify numerator: \[ Y(s) = \frac{64}{s^2(s^2 + 16)} + \frac{4s}{s^2 + 16} + \frac{5}{s^2 + 16} \] Now, take the inverse Laplace transform term by term: - \( \mathcal{L}^{-1}\left\{ \frac{64}{s^2(s^2 + 16)} \right\} \) gives \( 4t - 4\sin(4t) \) - \( \mathcal{L}^{-1}\left\{ \frac{4s}{s^2 + 16} \right\} \) gives \( 4\cos(4t) \) - \( \mathcal{L}^{-1}\left\{ \frac{5}{s^2 + 16} \right\} \) gives \( \frac{5}{4}\sin(4t) \) So, combine all terms: \[ y(t) = 4t + 4\cos(4t) - 4\sin(4t) + \frac{5}{4}\sin(4t) \] \[ y(t) = 4t + 4\cos(4t) + \left(-4 + \frac{5}{4}\right)\sin(4t) \] \[ y(t) = 4t + 4\cos(4t) - \frac{11}{4}\sin(4t) \] **Explanation:** In this step, we solved for \( Y(s) \), decomposed the terms, and found the inverse Laplace transform to obtain the final solution for \( y(t) \). --- ### **Final Answer:** \[ \boxed{ y(t) = 4t + 4\cos(4t) - \frac{11}{4}\sin(4t) } \] This is the solution to the initial value problem.

**Step 1: Take the Laplace transform of both sides** Given the differential equation: \[ y'' + 16 y = 64 t \] with initial conditions: \[ y(0) = 4, \quad y'(0) = 5 \] Applying the Laplace transform to both sides: \[ \mathcal{L}\{ y'' \} + 16 \mathcal{L}\{ y \} = \mathcal{L}\{ 64 t \} \] Recall: - \(\mathcal{L}\{ y''(t) \} = s^2 Y(s) - s y(0) - y'(0)\) - \(\mathcal{L}\{ y(t) \} = Y(s)\) - \(\mathcal{L}\{ 64 t \} = \frac{64}{s^2}\) Substituting: \[ s^2 Y(s) - s \cdot 4 - 5 + 16 Y(s) = \frac{64}{s^2} \] Simplify: \[ s^2 Y(s) + 16 Y(s) = \frac{64}{s^2} + 4s + 5 \] **Explanation:** In this step, we converted the differential equation into an algebraic equation in the Laplace domain, incorporating initial conditions. This sets the stage for solving for \(Y(s)\). --- **Step 2: Solve for \(Y(s)\) and find the inverse Laplace transform** Factor \(Y(s)\): \[ Y(s) (s^2 + 16) = \frac{64}{s^2} + 4s + 5 \] \[ Y(s) = \frac{\frac{64}{s^2} + 4s + 5}{s^2 + 16} \] Express as separate fractions: \[ Y(s) = \frac{64}{s^2 (s^2 + 16)} + \frac{4s}{s^2 + 16} + \frac{5}{s^2 + 16} \] Now, find inverse Laplace transforms of each term: 1. \(\mathcal{L}^{-1} \left\{ \frac{64}{s^2 (s^2 + 16)} \right\}\): This can be handled via partial fractions or known transforms, resulting in: \[ 4t - 4 \sin(4t) \] 2. \(\mathcal{L}^{-1} \left\{ \frac{4s}{s^2 + 16} \right\}\): Standard form gives: \[ 4 \cos(4t) \] 3. \(\mathcal{L}^{-1} \left\{ \frac{5}{s^2 + 16} \right\}\): Standard form gives: \[ \frac{5}{4} \sin(4t) \] **Final combined solution:** \[ \boxed{ y(t) = 4t + 4 \cos(4t) - \frac{11}{4} \sin(4t) } \] --- **Final answer:** \[ \boxed{ \boxed{ y(t) = 4t + 4 \cos(4t) - \frac{11}{4} \sin(4t) } } \]